PHY6426/Fall 2007: CLASSICAL MECHANICS
Homework #12: Solutions
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Due: Monday, Dec.3, 2007
Please help your instructor by doing your work neatly.
1. Goldstein, 13-4
L=
h2 ~
~ ∗ + V ψ ∗ ψ + h (ψ ∗ ψt − ψψt∗ ) .
∇ψ · ∇ψ
2
8π m
4πi
(1)
We treat ψ and ψ ∗ as two independent fields. The Euler’s equation for ψ ∗
d ∂L
d ∂L
∂L
+ i
=
,
∗
∗
dt ∂ψt
dx ∂ψ,i
∂ψ ∗
where ψ,i∗ ≡ dψ ∗ /dxi with i = 1 . . . 3 and summation over i is implied. Re-writing the first term in Eq.(1) as
∗
h2 dψ dψ
8π 2 m dxi dxi
and calculating the derivatives, we obtain
−
h
h2 d dψ
h
ψt + 2
=Vψ+
ψt
4πi
8π m dxi dxi
4πi
or
hi
h2
ψt = − 2 ∇2 ψ + V ψ.
2π
8π m
Same for ψ ∗ except for the sign of the left-hand side.
Momenta
∂L
h ∗
=
ψ
∂ψt
4πi
h
∂L
=−
=
ψ
∗
∂ψt
4πi
Πψ =
Πψ ∗
Hamiltonian density
∂L
∂L ∗
h ∗
h
h2 ~
∗
~ ∗ − V ψ ∗ ψ − h (ψ ∗ ψt − ψψt∗ )
ψt +
ψ
−
L
=
ψ
ψ
−
ψψ
−
∇ψ · ∇ψ
t
t
t
∂ψt
∂ψt∗
4πi
4πi
8π 2 m
4πi
2
h ~
~ ∗ − V ψ∗ψ
= − 2 ∇ψ
· ∇ψ
8π m
H =
2. Goldstein, 13-10.
The Lagrangian density depends on ψt , ψx , and ψxx . We need first to generalize the least action principle for
this case. Assuming, as in the class, that the trial field is ψ (x, t) = ψ0 (x, t) + βξ (x, t) and forming the action
as
Z
Z
I = dt dxL (ψt , ψx , ψxx ) ,
we impose the condition dI/dβ =
R
dt
R
dxdL (ψt , ψx , ψxx ) /dβ = 0. The derivative of L is
∂L
∂L
∂L
dL
=
ξt +
ξx +
ξxx .
dα
∂ψt
∂ψx
∂ψxx
2
Subsitute the dL
dβ and integrate by parts. In the last term, integrate by parts (with respect to x) twice. The
Euler equation is
−
d ∂L
d2 ∂L
d ∂L
−
+ 2
= 0.
dt ∂ψt
dx ∂ψx
dx ∂ψxx
Now,
∂L
1
= ψx
∂ψt
2
1
1
∂L
= ψt + αψx2
∂ψx
2
2
∂L
= −νψxx
∂ψxx
and the Euler equation reduces to
1
1
− ψxt − ψxt − αψx ψxx − ψxxxx = 0
2
2
which is the KdV equation for φ = ψx .
3. In addition to a single-soliton solution discussed in the class, the sine-Gordon equation has other solutions. In
particular,
sin (ut)
−1
φ1 (x, t) = 4 tan
uγ sinh (γ −1 x)
sinh (uγt)
−1
φ2 (x, t) = 4 tan
u sinh (γx)
sinh (γx)
−1
φ3 (x, t) = 4 tan
u
sinh (uγt)
are all supposed to be solutions of the sine-Gordon equation.
a) Verify that φ1,2,3 indeed satisfy the sine-Gordon equation.
Equation of motion
φxx = φtt + sin φ
(2)
For any solutions φ = 4 tan−1 [(F (x) g(t)] ≡ 4α. A little bit of trigonometry
sin 4α = 2 sin (2α) cos (2α) = 4 sin α cos α cos2 α − sin2 α = 4 sin α cos3 α 1 − tan2 α = 4 tan α cos4 α 1 − tan2 α
= 4 tan α
1 − tan2 α
1 − (F G)2
= 4F G 2
2
1 + tan2 α
1 + (F G)
Fx G
dφ3
= 4
2
dx
1 + (F G)

2
2Fx2 F G3

d φ3
4
 Fxx G

2
2
3
= 4
−
2  = 2 Fxx G 1 + (F G) − 2Fx F G
2
2
dx
2
2
1 + (F G)
1 + (F G)
1 + (F G)
F Gt
dφ3
= 4
2
dt
1 + (F G)


4F
2G2t GF 3 
d 2 φ3
 F Gtt
2
2
3
=
4
−
2G
GF
=
G
F
1
+
(F
G)
−


2
2
tt
t
2
dt2
1 + (F G)
1 + (F G)2
1 + (F G)2
3
Substuting these results into Eq.(2), we obtain
2
2
2
Fxx G 1 + (F G) − 2Fx2 F G3 − Gtt F 1 + (F G) − 2G2t GF 3 = F G 1 − (F G)
or, dividing by F G 1 + (F G)2
2
Fxx
2G2t F 2
1 − (F G)
Gtt
2Fx2 G2
+
=
−
−
2
2
2
F
G
1 + (F G)
1 + (F G)
1 + (F G)
(3)
For the solution φ3 it is convenient to introduce a function
G = 1/g
Gt = −gt /g 2
Gtt = −gtt /g 2 + 2gt2 /g 3
Gtt /G = −gtt /g + 2gt2 /g 2 .
Eq. (3b) is re-written as
gtt
g2
F2
g2 F 2
g2 − F 2
Fxx
+
− 2 t2 − 2 2 x 2 + 2 t2 2
= 2
2
F
g
g
F +g
g F +g
g + F2
or
gtt
g2
F2
g2 − F 2
Fxx
+
−2 2 t 2 −2 2 x 2 = 2
F
g
F +g
F +g
g + F2
For the solution φ3 , f = u sinh (γx) and G = sinh (uγt)
fx = uγ cosh (γx)
fxx = uγ 2 sinh (γx)
fxx /f = γ 2
gt = uγ cosh (uγt)
gtt = u2 γ 2 sinh (uγt)
gtt /g = u2 γ 2
sinh2 uγt + u2 sinh2 γx − 2u2 γ 2 cosh2 uγt − 2u2 γ 2 cosh2 ux = sinh2 (uγt) − u2 sinh2 ux
γ 2 + u2 γ 2 sinh2 uγt + u2 sinh2 γx − 2u2 γ 2 [2 + sinh2 uγt + sinh2 ux] = sinh2 (uγt) − u2 sinh2 ux
γ 2 + u2 γ 2
Equating the coefficients in front of sinh2 uγt and sinh2 γx, we see that the equation is satisfied if
γ2 =
1
.
1 − u2
Same for other solutions.
b) One of these solutions describes a soliton-soliton pair. Another one describes a solition-antisoliton pair.
The remaining one describes the localized perturbation (so-called ”breather”). Identify which one is which by
plotting the solutions.
φ3 : soliton-solition
φ2 : soliton-antisoliton
φ1 : breather